TOP 40 câu Trắc nghiệm Giới hạn của hàm số (có đáp án 2023) – Toán 11

Đáp án: C

Giải thích:

$\begin{array}{l}\underset{x\to 3}{\lim }\sqrt{\frac{9x^2-x}{(2x-1)(x^4-3)}}\\ =\sqrt{\frac{{9.3}^2-3}{(2.3-1)(3^4-3)}}\end{array}$

$=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

Đáp án: B

Giải thích:

Ta có $x<0$ nên:

$\begin{array}{l}\underset{x\to -\infty }{\lim }\left({\left|x\right|}^3+2x^2+3\left|x\right|\right)\\ =\underset{x\to -\infty }{\lim }\left(-x^3+2x^2-3x\right)\\ =\underset{x\to -\infty }{\lim }{x}^3.\left(-1+\frac{2}{x}-\frac{3}{x^2}\right)\\ =+\infty \end{array}$

Vì $\begin{array}{l}\underset{x\to -\infty }{\lim }{x}^3=-\infty ,\\ \underset{x\to -\infty }{\lim }\left(-1+\frac{2}{x}-\frac{3}{x^2}\right)=-1<0\end{array}$

Đáp án: C

Giải thích:

$\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x+3}-x\right)\\ =\underset{x\to +\infty }{\lim }\frac{x^2+x+3-x^2}{\sqrt{x^2+x+3}+x}\\ =\underset{x\to +\infty }{\lim }\frac{x+3}{\sqrt{x^2+x+3}+x}\\ =\underset{x\to +\infty }{\lim }\frac{1+\frac{3}{x}}{\sqrt{1+\frac{1}{x}+\frac{3}{x^2}}+1}\\ =\frac{1+0}{\sqrt{1+0+0}+1}=\frac{1}{2}\end{array}$

Đáp án: D

Giải thích:

$\begin{array}{l}\underset{x\to -\infty }{\lim }\left(\sqrt[3]{x^3+1}+x-1\right)\\ =\underset{x\to -\infty }{\lim }\left(x\sqrt[3]{1+\frac{1}{x^3}}+x-1\right)\\ =\underset{x\to -\infty }{\lim }\left[x\left(\sqrt[3]{1+\frac{1}{x^3}}+1-\frac{1}{x}\right)\right]\\ =-\infty \end{array}$

Vì $\underset{x\to -\infty }{\lim }x=-\infty ;$

$\hspace{0.17em}\underset{x\to -\infty }{\lim }\left(\sqrt[3]{1+\frac{1}{x^3}}+1-\frac{1}{x}\right)=2>\text{}0$

Đáp án: C

Giải thích:

Với $x\in \left(-1;0\right)$ thì $x+1>0$ và $\frac{x}{x-1}>0$

Do đó 

$\begin{array}{l}\underset{x\to {(-1)}^{+}}{\lim }\left(x^3+1\right)\sqrt{\frac{x}{x^2-1}}\\ =\underset{x\to {(-1)}^{+}}{\lim }(x+1)(x^2-x+1)\sqrt{\frac{x}{(x-1)(x+1)}}\\ =\underset{x\to {(-1)}^{+}}{\lim }\sqrt{x+1}(x^2-x+1)\sqrt{\frac{x}{x-1}}\\ =\sqrt{-1+\text{}1}.\text{\hspace{0.17em}[}-(-1)+\text{}1].\sqrt{\frac{-1}{-1-1}}=0\end{array}$

Đáp án: B

Giải thích:

$\begin{array}{l}\underset{x\to 2}{\lim }\sqrt[3]{\frac{x^2-x-1}{x^2+2x}}\\ =\sqrt[3]{\frac{2^2-2-1}{2^2+2.2}}\\ =\sqrt[3]{\frac{1}{8}}=\frac{1}{2}\end{array}$

Đáp án: B

Giải thích:

$\begin{array}{l}\underset{x\to \sqrt{3}}{\lim }\left|x^2-4\right|=\left|{\left(\sqrt{3}\right)}^2-4\right|\\ =\left|-1\right|=1\end{array}$

Đáp án: D

Giải thích:

$\begin{array}{l}\underset{x\to -\infty }{\lim }(x-x^3+1)\\ =\underset{x\to -\infty }{\lim }{x}^3\left(\frac{1}{x^2}-1+\frac{1}{x^3}\right)\\ =+\infty \end{array}$

vì $\left\{\begin{array}{l}\underset{x\to -\infty }{\lim }{x}^3=-\infty \\ \underset{x\to -\infty }{\lim \left(\frac{1}{x^2}-1+\frac{1}{x^3}\right)}=-1<0\end{array}\right.$

Đáp án: A

Giải thích:

Vì $\begin{array}{l}\left\{\begin{array}{l}\underset{x\to 2^{+}}{\lim }(x-15)=2-15=-13<0\\ \underset{x\to 2^{+}}{\lim }(x-2)=2-2=0\\ x-2>0,\forall x>2\end{array}\right.\\ \Rightarrow \underset{x\to 2^{+}}{\lim }\frac{x-15}{x-2}=-\infty \end{array}$

Đáp án: B

Giải thích:

Vì

$\left\{\begin{array}{l}\underset{x\to 2^{+}}{\lim }\sqrt{x+2}=\sqrt{2+\text{}2}=2>0\\ \underset{x\to 2^{+}}{\lim }\sqrt{x-2}=\sqrt{2-2}=0\\ \sqrt{x-2}>0,\forall x>2\end{array}\right.$

$\Rightarrow \underset{x\to 2^{+}}{\lim }\frac{\sqrt{x+2}}{\sqrt{x-2}}=+\infty$

Đáp án: B

Giải thích:

Ta có: 

$\underset{x\to +\infty }{\lim }f\left(x\right)=+\infty$

$\iff \underset{x\to +\infty }{\lim }\left[-f\left(x\right)\right]=-\infty$

Đáp án: B

Giải thích:

$\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+1}+x\right)$

$=\underset{x\to +\infty }{\lim }x\left(\sqrt{1+\frac{1}{x^2}}+1\right)$

$=+\infty$

vì $\left\{\begin{array}{l}\underset{x\to +\infty }{\lim }x=+\infty \\ \underset{x\to +\infty }{\lim }\left(\sqrt{1+\frac{1}{x^2}}+1\right)=2>0\end{array}\right.$

Đáp án: B

Giải thích:

$\underset{x\to +\infty }{\lim }\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+2}\right)$

$=\underset{x\to +\infty }{\lim }x\left(\sqrt[3]{3-\frac{1}{x^3}}+\sqrt{1+\frac{2}{x^2}}\right)$

$=+\infty$

Vì

$\left\{\begin{array}{l}\underset{x\to +\infty }{\lim }x=+\infty \\ \underset{x\to +\infty }{\lim }\left(\sqrt[3]{3-\frac{1}{x^3}}+\sqrt{1+\frac{2}{x^2}}\right)\\ =\sqrt[3]{3}+1>0\end{array}\right.$

Đáp án: B

Giải thích:

$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\sqrt{3x^2+1}$

$=\sqrt{{3.1}^2+1}=2$

Đáp án: C

Giải thích:

$\underset{x\to 1}{\lim }\frac{x-x^3}{(2x-1)(x^4-3)}$

$=\frac{1-1^3}{(2.1-1)(1^4-3)}=0$

Đáp án: C

Giải thích:

$\begin{array}{l}\underset{x\to 2}{\lim }\frac{x^3-6x^2+11x-6}{x^2-4}\\ =\underset{x\to 2}{\lim }\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)}\\ =\underset{x\to 2}{\lim }\frac{\left(x-1\right)\left(x-3\right)}{\left(x+2\right)}\\ =\frac{(2-1)(2-3)}{2+2}=\frac{-1}{4}\end{array}$

Đáp án: D

Giải thích:

$\begin{array}{l}\underset{x\to -1}{\lim }\frac{x^5+1}{x^3+1}\\ =\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\underset{x\to -1}{\lim }\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\end{array}$

Đáp án: B

Giải thích:

$\begin{array}{l}\underset{x\to 2}{\lim }\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\\ =\underset{x\to 2}{\lim }\frac{(x-\sqrt{x+2})(x+\sqrt{x+2})(\sqrt{4x+1}+3)}{(\sqrt{4x+1}-3)(\sqrt{4x+1}+3)(x+\sqrt{x+2})}\\ =\underset{x\to 2}{\lim }\frac{(x^2-x-2)(\sqrt{4x+1}+3)}{(4x+1-9)(x+\sqrt{x+2})}\\ =\underset{x\to 2}{\lim }\frac{(x+1).(x-2)(\sqrt{4x+1}+3)}{4.(x-2)(x+\sqrt{x+2})}\\ =\underset{x\to 2}{\lim }\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}\\ =\frac{(2+1)(\sqrt{4.2+1}+3)}{4.(2+\sqrt{2+2})}\\ =\frac{18}{16}=\frac{9}{8}\end{array}$

Đáp án: A

Giải thích:

$\begin{array}{l}\underset{x\to -\infty }{\lim }(x-1)\sqrt{\frac{x^2}{2x^4+x^2+1}}\\ =\underset{x\to -\infty }{\lim }\left[-\sqrt{\frac{x^2{(x-1)}^2}{2x^4+x^2+1}}\right]\\ =\underset{x\to -\infty }{\lim }\left[-\sqrt{\frac{x^2(x^2-2x+1)}{2x^4+x^2+1}}\right]\\ =\underset{x\to -\infty }{\lim }\left[-\sqrt{\frac{x^4-2x^3+x^2}{2x^4+x^2+1}}\right]\\ =\underset{x\to -\infty }{\lim }\left[-\sqrt{\frac{1-\frac{2}{x}+\frac{1}{x^2}}{2+\frac{1}{x^2}+\frac{1}{x^4}}}\right]\\ =\frac{-1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\end{array}$

Đáp án: C

Giải thích:

$\begin{array}{l}\underset{x\to -3}{\lim }\left|\frac{-x^2-x+6}{x^2+3x}\right|\\ =\underset{x\to -3}{\lim }\left|\frac{-(x+3)(x-2)}{x(x+3)}\right|\\ =\underset{x\to -3}{\lim }\left|\frac{-(x-2)}{x}\right|\\ =\left|\frac{-(-3-2)}{-3}\right|=\frac{5}{3}\end{array}$

Đáp án: D

Giải thích:

$\begin{array}{l}\underset{x\to 3}{\lim }\frac{x^2-4x+3}{x^2-9}\\ =\underset{x\to 3}{\lim }\frac{(x-1)(x-3)}{(x-3)(x+3)}\\ =\underset{x\to 3}{\lim }\frac{x-1}{x+3}=\frac{3-1}{3+3}\\ =\frac{1}{3}\end{array}$

Đáp án: B

Giải thích:

Ta có $3-x>0$ với mọi $x<3$, do đó:

$\begin{array}{l}\underset{x\to 3^{-}}{\lim }\frac{3-x}{\sqrt{27-x^3}}\\ =\underset{x\to 3^{-}}{\lim }\frac{3-x}{\sqrt{(3-x)(9+3x+x^2)}}\\ =\underset{x\to 3^{-}}{\lim }\frac{\sqrt{3-x}}{\sqrt{9+3x+x^2}}\\ =\frac{\sqrt{3-3}}{\sqrt{9+3.3+3^2}}=0\end{array}$

Đáp án: A

Giải thích:

Ta có 

$\begin{array}{l}\underset{x\to -1}{\lim }\frac{\sqrt{3x^2+1}-x}{x-1}\\ =\frac{\sqrt{3+1}+1}{-1-1}\\ =-\frac{3}{2}\end{array}$

Đáp án: D

Giải thích:

$\begin{array}{l}\underset{x\to -\infty }{\lim }\frac{3x^2-2x-1}{x^2+1}\\ =\underset{x\to -\infty }{\lim }\frac{3-\frac{2}{x}-\frac{1}{x^2}}{1+\frac{1}{x^2}}\\ =\frac{3-0-0}{1+0}=3\end{array}$

Đáp án: D

Giải thích:

$\begin{array}{l}\underset{x\to +\infty }{\lim }\frac{\sqrt{4x^2-2x+1}+2-x}{\sqrt{9x^2-3x}+2x}\\ =\underset{x\to +\infty }{\lim }\frac{x\sqrt{4-\frac{2}{x}+\frac{1}{x^2}}+\text{}2-x}{x.\sqrt{9-\frac{3}{x}}+2x}\\ =\underset{x\to +\infty }{\lim }\frac{\sqrt{4-\frac{2}{x}+\frac{1}{x^2}}+\frac{2}{x}-1}{\sqrt{9-\frac{3}{x}}+2}\\ =\frac{\sqrt{4-0+0}+\text{}0-1}{\sqrt{9-0}+\text{}2}\\ =\frac{1}{5}\end{array}$

Đáp án: B

Giải thích:

$\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+3x}-\sqrt{x^2+4x}\right)\\ =\underset{x\to +\infty }{\lim }\frac{(\sqrt{x^2+3x}-\sqrt{x^2+4x}).(\sqrt{x^2+3x}+\sqrt{x^2+4x})}{\sqrt{x^2+3x}+\sqrt{x^2+4x}}\\ =\underset{x\to +\infty }{\lim }\frac{x^2+3x-x^2-4x}{x\sqrt{1+\text{\hspace{0.17em}}\frac{3}{x}}+x\sqrt{1+\frac{4}{x}}}\\ =\underset{x\to +\infty }{\lim }\frac{-x}{x.(\sqrt{1+\text{\hspace{0.17em}}\frac{3}{x}}+\sqrt{1+\frac{4}{x}})}\\ =\underset{x\to +\infty }{\lim }\frac{-1}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{4}{x}}}\\ =-\frac{1}{2}\end{array}$

Đáp án: C

Giải thích:

Khi $x\to 3^{+}$ thì x > 3 nên x -  3> 0

$\Rightarrow \left|x-3\right|=x-3$

$\begin{array}{l}\underset{x\to 3^{+}}{\lim }\frac{\left|x-3\right|}{3x-9}=\underset{x\to 3^{+}}{\lim }\frac{x-3}{3x-9}\\ =\underset{x\to 3^{+}}{\lim }\frac{x-3}{3(x-3)}=\frac{1}{3}\end{array}$

Đáp án: D

Giải thích:

$\begin{array}{l}\underset{x\to -1^{+}}{\lim }\frac{x^2+3x+2}{\left|x+1\right|}\\ =\underset{x\to -1^{+}}{\lim }\frac{(x+1)(x+2)}{x+1}\\ =\underset{x\to -1^{+}}{\lim }(x+2)=-1+2=1\\ \underset{x\to -1^{-}}{\lim }\frac{x^2+3x+2}{\left|x+1\right|}\\ =\underset{x\to -1^{-}}{\lim }\frac{(x+1)(x+2)}{-(x+1)}\\ =\underset{x\to -1^{-}}{\lim }\left[-(x+2)\right]\\ =-(-1+2)=-1\\ \Rightarrow \underset{x\to -1^{+}}{\lim }\frac{x^2+3x+2}{\left|x+1\right|}\ne \underset{x\to -1^{-}}{\lim }\frac{x^2+3x+2}{\left|x+1\right|}\end{array}$

Suy ra, không tồn tại $\underset{x\to -1^{-}}{\lim }\frac{x^2+3x+2}{\left|x+1\right|}$.

Đáp án: C

Giải thích:

Ta có 

$\begin{array}{l}\underset{x\to 1}{\lim }\left(\frac{a}{1-x}-\frac{b}{1-x^3}\right)\\ =\underset{x\to 1}{\lim }\frac{a(1+x+x^2)-b}{(1-x).(1+x+x^2)}\\ =\underset{x\to 1}{\lim }\frac{a+ax+a{x}^2-b}{(1-x).(1+x+x^2)}\end{array}$

 Khi đó $\underset{x\to 1}{\lim }\left(\frac{a}{1-x}-\frac{b}{1-x^3}\right)$ hữu hạn thì tử thức $a+ax+a{x}^2-b$ phải có nghiệm bằng 1

⇔ $a+a.1+a{.1}^2-b=0$

$\iff 3a-b=0.$

Vậy ta có $\left\{\begin{array}{l}a+b=4\\ 3a-b=0\end{array}\right.\iff \left\{\begin{array}{l}a=1\\ b=3\end{array}\right.$

⇒L= $-\underset{x\to 1}{\lim }\left(\frac{a}{1-x}-\frac{b}{1-x^3}\right)$

$=-\underset{x\to 1}{\lim }\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$

$\begin{array}{l}=-\underset{x\to 1}{\lim }\frac{x^2+x-2}{\left(1-x\right).\left(1+x+x^2\right)}\\ =-\underset{x\to 1}{\lim }\frac{(x-1).(x+2)}{\left(1-x\right).\left(1+x+x^2\right)}\\ =-\underset{x\to 1}{\lim }\frac{-\left(x+2\right)}{1+x+x^2}=1\end{array}$

Đáp án: D

Giải thích:

Ta có:

$\begin{array}{l}\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1\\ =\sqrt{1+2x}-\sqrt{1+2x}\\ +\sqrt{1+2x}.\sqrt[3]{1+3x}\\ -\sqrt{1+2x}.\sqrt[3]{1+3x}\\ +\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1\\ =\left(\sqrt{1+2x}-1\right)\\ +\sqrt{1+2x}\left(\sqrt[3]{1+3x}-1\right)\\ +\sqrt{1+2x}.\sqrt[3]{1+3x}.\left(\sqrt[4]{1+4x}-1\right)\\ \Rightarrow \underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}\\ =\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+2x}-1}{x}\right)\\ +\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\frac{\sqrt[3]{1+3x}-1}{x}\right)\\ +\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\sqrt[4]{1+4x}-1}{x}\right)\end{array}$

Tính:

$\begin{array}{l}\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+2x}-1}{x}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{\left(\sqrt{1+2x}-1\right)\left(\sqrt{1+2x}+1\right)}{x\left(\sqrt{1+2x}+1\right)}\right)\\ =\underset{x\to 0}{\lim }\frac{2x}{x\left(\sqrt{1+2x}+1\right)}\\ =\underset{x\to 0}{\lim }\frac{2}{\sqrt{1+2x}+1}\\ =\frac{2}{1+1}=1\end{array}$

$\begin{array}{l}\underset{x\to 0}{\lim }\left(\sqrt{1+2x.}\frac{\left(\sqrt[3]{1+3x}-1\right)}{x}\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt{1+2x.}\frac{\left(\sqrt[3]{1+3x}-1\right)\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}{x\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt{1+2x.}\frac{3x}{x\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{3\sqrt{1+2x}}{\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)=\frac{3.1}{1+1+1}=1\end{array}$

$\begin{array}{l}\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\left(\sqrt[4]{1+4x}-1\right)}{x}\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\left(\sqrt[4]{1+4x}-1\right)\left[{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1\right]}{x\left[{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1\right]}\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{4x}{x\left[{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1\right]}\right)\\ =\underset{x\to 0}{\lim }\frac{4\sqrt{1+2x}.\sqrt[3]{1+3x}.}{{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1}\\ =\frac{\mathrm{4.1.1}}{1+1+1+1}=1\end{array}$

Vậy $\underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}$

$=1+1+1=3$

Đáp án: B

Giải thích: